A vector of magnitude 2 along a bisector of the angle between the two vectors 2→i−2→j+→k and →i+2→j−2→k is
A
2√10(3→i−→k)
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B
1√26(→i−4→j+3→k)
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C
2√26(→i−4→j+3→k)
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D
None of these
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Solution
The correct options are C2√26(→i−4→j+3→k) D2√10(3→i−→k) The vector along the angle bisector is obtained by the sum and difference of unit vectors along both arms.
Thus, 13(2→i−2→j+→k±→i+2→j−2→k)=→i−13→k and 13→i−43→j+→k
Now for the vector of magnitude 2, we multiply the unit vector in that direction by 2 and get 2√10(3→i−→k) and 2√26(→i−4→j+3→k)